Discrete Logging hunnu10590 pku2417 fzu 1352 hit 1928 zoj 1898

以下转自：http://hi.baidu.com/aekdycoin/blog/item/b317ca18bb24334942a9ad55.html

【普通Baby Step Giant Step】

【问题模型】
求解
A^x = B (mod C) 中 0 <= x < C 的解,C 为素数

【思路】
我们可以做一个等价
x = i * m + j  ( 0 <= i < m, 0 <=j < m) m = Ceil ( sqrt( C) )
而这么分解的目的无非是为了转化为:
(A^i)^m * A^j = B ( mod C)

之后做少许暴力的工作就可以解决问题：
(1) for i = 0 -> m, 插入Hash (i, A^i mod C)
(2) 枚举 i ,对于每一个枚举到的i,令  AA = (A^m)^i mod C
我们有
AA * A^j = B (mod C)
显然AA,B,C均已知，而由于C为素数,那么(AA,C)无条件为1
于是对于这个模方程解的个数唯一(可以利用扩展欧几里得或 欧拉定理来求解)
那么对于得到的唯一解X,在Hash表中寻找,如果找到，则返回 i * m + j
注意:由于i从小到大的枚举,而Hash表中存在的j必然是对于某个剩余系内的元素X 是最小的(就是指标)
所以显然此时就可以得到最小解

如果需要得到 x > 0的解，那么只需要在上面的步骤中判断 当 i * m + j > 0 的时候才返回


到目前为止，以上的算法都不存在争议,大家实现的代码均相差不大。可见当C为素数的时候，此类离散对数的问题可以变得十分容易实现。

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#define nmax 46341
#define LL long long
typedef struct Num {
	int nnum;
	int ii;
} Num;
Num num[nmax];
int x, y;
int cmp(const void *a, const void *b) {
	Num *n = (Num *) a;
	Num *m = (Num *) b;
	if (n->nnum > m->nnum) {
		return 1;
	}
	return -1;
}
int extend_gcd(int a, int b) {
	if (b == 0) {
		x = 1, y = 0;
		return a;
	}
	int d = extend_gcd(b, a % b);
	int tx = x;
	x = y;
	y = tx - a / b * y;
	return d;
}
int find_num(int x, int n) {
	int mid, left, right;
	left = 0, right = n + 1;
	while (left <= right) {
		mid = (left + right) >> 1;
		if (num[mid].nnum == x) {
			return num[mid].ii;
		} else if (num[mid].nnum > x) {
			right = mid - 1;
		} else {
			left = mid + 1;
		}
	}
	return -1;
}
int main() {
#ifndef ONLINE_JUDGE
	freopen("t.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif
	LL ptemp, te;
	int i, j, pte, p, b, n, bb, temp;
	while (scanf("%d %d %d", &p, &b, &n) != EOF) {
		pte = (int) ((sqrt(p * 1.0) + 0.5));
		for (i = 0, ptemp = 1; i <= pte; i++) {
			num[i].nnum = (int) (ptemp);
			num[i].ii = i;
			ptemp = ptemp * b % p;
		}
		bb = num[pte].nnum;
		qsort(num, pte + 1, sizeof(num[0]), cmp);
		for (i = 0, ptemp = 1; i <= pte; i++) {
			temp = (int) (ptemp);
			extend_gcd(temp, p);
			te = (int) (x);
			te = te * n;
			te = te % p + p;
			x = (int) (te % p);
			j = find_num(x, pte);
			if (j != -1) {
				printf("%d\n", pte * i + j);
				break;
			}
			ptemp = ptemp * bb % p;

		}
		if (i > pte) {
			printf("ERROR\n");
		}
	}
	return 0;
}

 

 

 hunnu 10590 fzu 1352 pku 2417 hit 1928 zoj 1898

/*
 a^x=b (mod c) c is prime
 */
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#define LL long long
#define nmax 46345
typedef struct num {
    int ii, value;
} num;
num Num[nmax];
int x, y;
int cmp(const void *a, const void *b) {
    num n = *(num *) a;
    num m = *(num *) b;
    return n.value - m.value;
}
void extend_gcd(int a, int b) {
    int xx;
    if (b == 0) {
        x = 1, y = 0;
        return;
    }
    extend_gcd(b, a % b);
    xx = x;
    x = y, y = xx - a / b * y;
}
int bfindNum(int key, int n) {
    int left, right, mid;
    left = 0, right = n;
    while (left <= right) {
        mid = (left + right) >> 1;
        if (Num[mid].value == key) {
            return Num[mid].ii;
        } else if (Num[mid].value > key) {
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    return -1;
}
void solve(int c, int a, int b) {
    int i, j, te, aa;
    LL temp, xx;
    te = (int) (sqrt(c * 1.0) + 0.5);
    for (i = 0, temp = 1 % c; i <= te; i++) {
        Num[i].ii = i;
        Num[i].value = (int) (temp);
        temp = temp * a % c;
    }
    aa = Num[te].value;
    qsort(Num, te + 1, sizeof(Num[0]), cmp);
    for (i = 0, temp = 1; i <= te; i++) {
        extend_gcd((int) (temp), c);
        xx = (LL) x;
        xx = xx * b;
        xx = xx % c + c;
        x = (int) (xx % c);
        j = bfindNum(x, te + 1);
        if (j != -1) {
            printf("%d\n", i * te + j);
            return;
        }
        temp = temp * aa % c;
    }
    puts("no solution");
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int p, b, n;
    while (~scanf("%d %d %d", &p, &b, &n)) {
        solve(p, b, n);
    }
    return 0;
}